F. Hitting the target
time limit per test
1.0 s
memory limit per test
512 MB
input
standard input
output
standard output
Sitting in front of the computer for too long is no good for your health. Keeping this in mind, the MaratonIME coaches decided to motivate the contestants to practice sports, not just programming. They didn't want to stick with the basics, so they chose the bow and arrow.
However, they noticed that too many people showed up at practice sessions just to shoot arrows. To discourage people with no interest in programming, they decided to add a requirement for shooting arrows: one would have to code a program that takes the coordinates of 3 arrows and record the final score. The target is made of ten concentric circles of proportional radii, that is, the smallest circle has radius R, the second one has radius 2R, the third one has radius 3R, and so on. The circle with smallest radius grants 10 points, and the largest one awards only 1 point. An arrow that hits the boundary between two circles grants the largest of the two scores.
Don't miss your opportunity to participate at the programming contest: write this program.
Input
The input has a line that has an integer R, the radius of the smallest circle. Next there are three lines with pairs of integers, x and y, the points in the target hit by each arrow. The target center is at coordinate (0, 0).
- 1 ≤ R ≤ 102
- - 103 ≤ x, y ≤ 103
Output
The output has a single integer: the final score.
Examples
input
Copy
1
0 0
0 1
0 0
output
Copy
30
input
Copy
2
0 0
0 2
0 3
output
Copy
29
此题纯水题,但是我还是CE了很多次;
就是把r=sqrt(double(x*x+y*y));写成了r=sqrt(x*x+y*y);
AC代码:
#include <iostream> #include <cstdio> #include <algorithm> #include <math.h> #include <cmath> #include <cstring> using namespace std; int main(){ int R; int x,y; int point; int sum=0; double r; cin>>R; for(int i=1;i<=3;i++){ cin>>x>>y; r=sqrt(double(x*x+y*y)); if(r>=0&&r<=R) point=10; else if(r>R&&r<=2*R) point=9; else if(r>2*R&&r<=3*R) point=8; else if(r>3*R&&r<=4*R) point=7; else if(r>4*R&&r<=5*R) point=6; else if(r>5*R&&r<=6*R) point=5; else if(r>6*R&&r<=7*R) point=4; else if(r>7*R&&r<=8*R) point=3; else if(r>8*R&&r<=9*R) point=2; else if(r>9*R&&r<=10*R) point=1; else if(r>10*R) point=0; sum+=point; } printf("%d ",sum); return 0; }
CE代码:
#include <iostream> #include <cstdio> #include <algorithm> #include <math.h> #include <cmath> #include <cstring> using namespace std; int main(){ int R; int x,y; int point; int sum=0; double r; cin>>R; for(int i=1;i<=3;i++){ cin>>x>>y; r=sqrt(x*x+y*y); if(r>=0&&r<=R) point=10; else if(r>R&&r<=2*R) point=9; else if(r>2*R&&r<=3*R) point=8; else if(r>3*R&&r<=4*R) point=7; else if(r>4*R&&r<=5*R) point=6; else if(r>5*R&&r<=6*R) point=5; else if(r>6*R&&r<=7*R) point=4; else if(r>7*R&&r<=8*R) point=3; else if(r>8*R&&r<=9*R) point=2; else if(r>9*R&&r<=10*R) point=1; else if(r>10*R) point=0; sum+=point; } printf("%d ",sum); return 0; }
CE代码提交结果:
"Canu0027t compile file: program.cpp program.cpp(17) : error C2668: u0027sqrtu0027 : ambiguous call to overloaded function c:Program Files (x86)Microsoft Visual Studio 10.0VCINCLUDEmath.h(589): could be u0027long double sqrt(long double)u0027 c:Program Files (x86)Microsoft Visual Studio 10.0VCINCLUDEmath.h(541): or u0027float sqrt(float)u0027 c:Program Files (x86)Microsoft Visual Studio 10.0VCINCLUDEmath.h(127): or u0027double sqrt(double)u0027 while trying to match the argument list u0027(int)u0027 "